You have found the following ages (in years) of all 4 seals at your local zoo: $ 9,\enspace 18,\enspace 9,\enspace 9$ What is the average age of the seals at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{9 + 18 + 9 + 9}{{4}} = {11.3\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $9$ years $-2.3$ years $5.29$ years $^2$ $18$ years $6.7$ years $44.89$ years $^2$ $9$ years $-2.3$ years $5.29$ years $^2$ $9$ years $-2.3$ years $5.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{5.29} + {44.89} + {5.29} + {5.29}} {{4}} $ $ {\sigma^2} = \dfrac{{60.76}}{{4}} = {15.19\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{15.19\text{ years}^2}} = {3.9\text{ years}} $ The average seal at the zoo is 11.3 years old. There is a standard deviation of 3.9 years.